Thread: UFDs, primes and irreducibles - Dummit and Foote - Chapter 8 - Proposition 11, page 2

I am reading Dummit and Foote Ch 8, Section 8.3 Unique Factorization Domains (UFDs)

I am studying Proposition 11 on page 284 (see attachment) which reads:

"In a Principal Ideal Domain, a non-zero element is a prime if and only if it is irreducible"

I follow the proof (see attachment) down to the statement:

But p is irreducible so by definition either r or m is a unit (OK so far!)

[ so we have if r is a unit then there exists an element u of the PID such that ru = ur = 1 and if m is a unit then there exist a PID element v such that mv = vm = 1]


But then D&F state the folowing:

"This means that either (p) = (m) or (m) = (1) respectively."

Can anyone show me formally how this follows from the statement that either r or m is a unit?

WOuld appreciate the help!

Peter

They're implicitly using that (x) = (ux) for all units u in R.
Have (p) contained in an ideal I implies (via PID) that I = (m) for some m.
Now p in (m), so there exists r such that p = rm.
By p irreducible, p = rm implies that either r or m is a unit.
If r is a unit, then from p = rm, have (p) = (rm) = (m) = I. Thus (p) = I.
If m is a unit, then I = (m) = (m*m) = (1) = R. Thus I = R. (Where m* is the multiplicative inverse of m, which is itself a unit.)
Thus (p) is a maximal ideal, because the only ideals containing (p) are (p) itself and R.
Maximal ideals are prime, thus (p) is a prime ideal.
Thus p is a prime element.

Thanks for the help

Can you demonstrate that (x) = (ux) for all units u in R

Will work on this myself as well

it suffices to show that x is in (ux), and ux is in (x).

that ux is in (x) is trivial.

this means that (ux) is contained in (x).

on the other hand, x = 1x = (u^-1u)x = u^-1(ux), which is clearly in (ux).

thus (x) is contained in (ux), as well, and the two sets are equal.

in a PID, ideal containment is equivalent to the notion of divisibility:

(y) is contained in (x) iff x divides y.

for example, in the integers, (6) is contained in (2), because 2 divides 6.

in other words: in order to get the equivalent concept of "prime numbers" in a PID, what we want to do is work with "prime ideals".

this lets us generalize the following result:

Z_n is an integral domain iff n is a prime number, to:

R/P is an integral domain iff P is a prime ideal.

this is a handy way to get rid of unwanted zero-divisors and make a "nicer" ring from an "ugly" one.

in PID's, non-zero prime ideals are also maximal, so for a PID D, and a prime ideal P, D/P is actually a field (the very nicest kind of ring).

a special case of PID's is where R = F[x], the ring of polynomials over a field F. the "prime elements" in such a ring are the irreducible polynomials.

this gives a very nice way of constructing the complex numbers:

C = R[x]/(x²+1), the coset x + (x²+1) plays the role of the complex number i (all higher degree polynomials are absorbed into the ideal, except for a linear factor remainder). this is really just a fancy way of saying:

C = R[√(-1)] = R(i).

one advantage to this (as opposed to say defining C as a 2-dimensional vector space over R with a multiplication), is that we immediately get all the field axioms "for free", from properties of RINGS.

Thanks Deveno ... Your detailed post is very helpful ... And much appreciated

Peter